Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I want to find some subset of R^2 which its intersection with every vertical line is measure zero if we see it as a subset of R and it is not measure zero in R^2?

share|improve this question
Why?${}{}{}{}{}$ –  Asaf Karagila Mar 14 '14 at 11:27
because of some reasons in integrability of multivariable functions –  alich Mar 14 '14 at 11:29
Wouldn't this contradict Fubini's theorem? Of course, assuming that the set is measurable in the first place. –  Alex Degtyarev Mar 14 '14 at 11:31
Why does It contradict Fubini theorem? –  alich Mar 14 '14 at 11:35
@alich: Because such a set $A\subseteq\mathbb{R}^2$ would satisfy $0\neq \lambda^2(A) = \int_{\mathbb{R}^2} \chi_A(x,y) d\lambda^2(x,y) = \int_\mathbb{R} \int_\mathbb{R} \chi_A(x,y) d\lambda^1(x) d\lambda^1(y) = \int_\mathbb{R} 0 d\lambda^1(y) = 0$. (where $\lambda^{1,2}$ denotes the 1-dimensional and 2-dimensional lebesgue meausure respectively) –  Johannes Hahn Mar 14 '14 at 11:59

1 Answer 1

up vote 9 down vote accepted

It is a well known result of Sierpinski that there exists non-measurable subsets of $\mathbb{R}^{2}$ which intersect each line in at most two points. Furthermore, there exists a real-valued function whose graph is a non-measurable subset of $\mathbb{R}^{2}$.

  1. W. Sierpi´nski: Sur un probl`eme concernant les ensembles mesurables superficiellement. Fund. Math. 1 (1920), 112–115.

  2. Gelbaum, Bernard R., and John M. H. Olmsted. Counterexamples in Analysis. San Francisco: Holden-Day, 1964. (p. 142-145)

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.